Chapter 3: Fourier Series of periodic signals
ECSE-303A Signals and Systems I Wednesday, February 11, 2009
LECTURE 12 Sections 3.1-3.3 Introduction to the Fourier series of periodic signals
Chapter 3: Fourier Series of periodic signals 3.0 3.1 3.2 3.3 3.4 3.5 3.8 3.9
Introduction Historical perspective The response of LTI systems to complex exponentials Fourier series representation of continuous-time periodic signals Convergence of Fourier series Properties of continuous-time Fourier series Fourier series and LTI systems Filtering
Q. How to make a rectangular shape out of circles?
2/11/2009 ECSE-303A
The Fourier series
By analogy, could we plot a square function using sinusoidal functions? f(t) t
In the 1800’s, Jean Baptiste Joseph Fourier found that most periodic functions with finite average power could be represented by a sum of sines and cosines
x(t ) =
+∞
∑
k =−∞
=
ak e jkω0t
+∞
∑a
k =−∞
k
⎡⎣ cos ( kω0t ) + j sin ( kω0t ) ⎤⎦
+∞
= a0 + ∑ Ak cos ( kω0t ) + Bk sin ( kω0t ) 2/11/2009 ECSE-303A
k =0
J.B. Fourier 1768-1830
Example
,
2/11/2009 ECSE-303A
f0 = 1Hz
+
f0 = 1Hz
f0 = 3Hz
= + f0 = 5Hz
2/11/2009 ECSE-303A
+ ... f0 = 7, 9...Hz
With three sin(2πf0t) components f0 = 1, 3, 5 Hz
With twentyfive sin(2πf0t) components f0 = 1, 3, 5...51 Hz 2/11/2009 ECSE-303A
To represent any arbitrary periodic function, what frequency component must be chosen? What is the amplitude coefficient of each term?
2/11/2009 ECSE-303A
Chapter 3: Fourier Series of periodic signals 3.0 Introduction 3.1 Historical perspective 3.2 The response of LTI systems to complex exponentials 3.3 Fourier series representation of continuous-time periodic signals 3.4 Convergence of Fourier series 3.5 Properties of continuous-time Fourier series 3.8 Fourier series and LTI systems 3.9 Filtering
Eigenfunctions of LTI Difference and Differential Systems We already know that the complex exponentials of the type Aest and Czn remain basically invariant under the action of time shifts (difference systems) and derivatives (differential systems). The response of an LTI system to a complex exponential input is the same complex exponential with only a change in (complex) amplitude. Continuous-time LTI system:
e → H ( s )e
Discrete-time LTI system:
z n → H ( z) z n
st
st
Where the complex amplitude factors H(s), H(z) are functions of the complex variable s or z.
2/11/2009 ECSE-303A
10
Input signals like x(t)=est and x[n]=zn for which the system output is a constant times the input signal are called eigenfunctions of the LTI system, and the complex gains are the system's eigenvalues corresponding to the eigenfunctions.
2/11/2009 ECSE-303A
11
To show that x(t)=est is indeed an eigenfunction of any LTI system of impulse response h(t), we look at the following convolution integral:
y (t ) =
+∞
∫ h (τ ) x ( t − τ ) dτ
−∞ +∞
=
∫
h (τ ) e s( t −τ ) dτ
−∞ +∞
= e st
∫
h (τ ) e − sτ dτ
−∞
The system's response has the form y(t)=H(s)est, where
z
+∞
H (s) = h(τ )e − sτ dτ is an eigenvalue and 2/11/2009 ECSE-303A
est
−∞
is an eigenfunction 12
Similarly for LTI discrete-time systems, the complex exponential x[n]=zn is an eigenfunction: +∞
y [ n] =
=
∑ h [k ] x [n − k ]
k =−∞ +∞
∑
k =−∞
= zn
h [ k ] z n−k
+∞
∑ h [k ] z
−k
k =−∞
The system's response has the form
y[n] = H ( z ) z n Where
H ( z) =
+∞
∑
h[k ]z − k
k =−∞
2/11/2009 ECSE-303A
13
The eigenvalues of LTI systems in continuous and discrete-time +∞
H ( s) =
∫
h(τ )e− sτ dτ
−∞
H ( z) =
+∞
∑ h[k ]z
−k
k =−∞
Are respectively known as the Laplace transform and the ztransform of the system's impulse response (also called transfer function, more of this later in the course and Signals 2).
2/11/2009 ECSE-303A
14
Chapter 3: Fourier Series of periodic signals 3.0 3.1 3.2 3.3 3.4 3.5 3.8 3.9
Introduction Historical perspective The response of LTI systems to complex exponentials Fourier series representation of continuous-time periodic signals Convergence of Fourier series Properties of continuous-time Fourier series Fourier series and LTI systems Filtering
Linear Combinations of HarmonicallyRelated Complex Exponentials Periodic signals satisfy
x (t ) = x (t + T )
{−∞ < t < +∞}
for some positive value of T. The smallest such T is the fundamental period and ω0=2π/T (radians/s) is the fundamental frequency A periodic signal x(t) is completely known from prior knowledge of only one period T A special case of periodic signals, the harmonically-related complex exponentials have frequencies that are integer multiples of ω0 jkω0t
φk (t ) e
2/11/2009 ECSE-303A
, k = 0,± 1,± 2,…
16
Harmonically related exponentials form an orthogonal set. That is, 2π
2π
ω0
ω0
0
0
* φ ( t ) φ ∫ k m (t )dt =
∫
e jkω0t e − jmω0t dt
2π
=
ω0
∫
e j ( k − m )ω0t dt
0
1 ⎡⎣e j ( k − m )2π − 1⎤⎦ = j (k − m)ω0 ⎧ 2π ⎪ , m=k = ⎨ ω0 ⎪ 0, m ≠ k ⎩ 2/11/2009 ECSE-303A
Use l’Hopital’s rule 17
A linear combination (i.e. summation) of harmonically-related complex exponentials φk(t)
x(t ) =
+∞
∑
k =−∞
akφk =
+∞
∑
k =−∞
ak e jkω0t =
+∞
∑ae
k =−∞
jk (
2π )t T
k
is also periodic with fundamental period T. Terms with k=±1 ⇒ The fundamental frequency component or the first harmonic of the signal Terms with k=±2 ⇒ second harmonic components (at frequency 2ω0) Terms with k=±N ⇒ Nth harmonic components.
The representation of periodic signals in the form given above is referred to as the Fourier series representation 2/11/2009 ECSE-303A
18
Example 4.1: Consider the periodic signal with fundamental frequency ω0=π/2 rad/s made up of the sum of five harmonic components:
x(t ) =
5
∑a e
k =−5
π
jk t 2
k
,
a0 = 0 a±1 = −0.2026 a±2 = 0 a±3 = −0.0225 a±4 = 0 a±5 = −0.0081 2/11/2009 ECSE-303A
19
Collecting the harmonic components together, we obtain
x(t ) =
5
∑a e
k =−5
π
jk t 2
k
= −0.2026(e
π
j t 2
+e
π
−j t 2
) − 0.0225(e
j
3π t 2
+e
−j
3π t 2
) − 0.0081(e
j
5π t 2
+e
−j
5π t 2
π
3π 5π = −0.4052 cos( t ) − 0.0450 cos( t ) − 0.0162 cos( t ) 2 2 2
2/11/2009 ECSE-303A
20
)
Relative amplitude
π 3π 5π x(t ) = −0.4052 cos( t ) − 0.0450 cos( t ) − 0.0162 cos( t ) 2 2 2
2/11/2009 ECSE-303A
Time [s]
21
Given a periodic function, how to represent it with a summation of harmonic components?
x(t ) =
+∞
∑ae
k =−∞
jkω0t
k
We use the orthogonality property of the harmonically related complex exponentials. Multiplying both sides of the above equation with another complex exponential and integrating over one fundamental period of the signal +∞
T
T
0
0 k =−∞
− jnω0t x ( t ) e dt = ∫ ∫
=
+∞
T
∑
∑ a ∫e
k =−∞
k
0
= Tak 2/11/2009 ECSE-303A
ak e jkω0t e − jnω0t dt T
jkω0t − jnω0 t
e
{k = n}
dt
1 − jnω0t ⇒ ak = ∫ x ( t ) e dt T 0 22
In conclusion, if a periodic signal x(t) has a Fourier series representation, then we have the Fourier series equation pair
x (t ) =
+∞
∑ae
k =−∞
k
jkω0t
=
+∞
∑ae
k =−∞
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
k
1 1 − jkω0t ak = ∫ x ( t ) e dt = ∫ x ( t ) e TT TT
2/11/2009 ECSE-303A
⎛ 2π − jk ⎜ ⎝ T
⎞ ⎟t ⎠
dt
23
x (t ) =
+∞
∑
k =−∞
ak e jkω0t =
+∞
∑ae
k =−∞
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
k
The first equation provides a time-domain representation of the signal as a sum of periodic complex exponential signals. This is the synthesis equation.
1 1 − jkω0t ak = ∫ x ( t ) e dt = ∫ x ( t ) e TT TT
⎛ 2π − jk ⎜ ⎝ T
⎞ ⎟t ⎠
dt
The second equation provides a frequency-domain representation of the signal. The ak represent the Fourier series coefficients, or the spectral coefficients of x(t). 2/11/2009 ECSE-303A
24
Example: The periodic “sawtooth” signal
The fundamental period is T=1s hence
ω0 =
2π = 2π rad/s T
The first ak coefficient with k=0.
1 1 − jkω0t ak = ∫ x( t )e dt ⇒ a0 = ∫ x( t )dt TT TT Since the average over one period is 0, we find a0=0. 2/11/2009 ECSE-303A
25
1 ak = ∫ x(t )e − jkω0t dt TT 1
= ∫ x(t )e − jk 2π t dt 0
1
= ∫ (1 − 2t )e − jk 2π t dt 0
− jk 2π t
1
− jk 2π t
1
⎡ ⎤ ⎡e ⎤ e = ⎢(1 − 2t ) −⎢ ( −2 ) ⎥ ⎥ − jk 2π ⎦ 0 ⎣ − jk 2π ⎦ 0 ⎣ 1 1 1 = + = jk 2π jk 2π jkπ −j = {k ≠ 0} kπ 2/11/2009 ECSE-303A
26
x (t ) =
+∞
∑ae
k =−∞
k = ±1
k
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
−j ⎧ ⎫ , k ≠ 0⎬ ⎨T = 1, ak = kπ ⎩ ⎭
k = ±1,2
k = ±1,2,3
2/11/2009 ECSE-303A
27
x (t ) =
+∞
∑ae
k =−∞
k
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
−j⎫ ⎧ ⎨T = 1, ak = ⎬ kπ ⎭ ⎩
k = ±∞
2/11/2009 ECSE-303A
28
ECSE-303A Signals and Systems I Wednesday, February 11, 2009
LECTURE 13 Section 3.3-More examples of Fourier series Section 3.4-Convergence of the Fourier series
Example Consider the following periodic rectangular wave of 2π fundamental period T and fundamental frequency ω 0 =
T
.
2/11/2009 ECSE-303A
30
The Fourier series coefficients are, for k=0 t0
2t0 1 a0 = ∫ 1dt = T −t0 T
a0: average value of x(t)
x (t ) =
+∞
∑ae
k =−∞
ak =
jkω0 t
k
1 − jkω0 t x t e dt ( ) ∫ TT
and for k≠0 t
1 0 − jkω0t ak = ∫ e dt T − t0 1
=−
⎡⎣e jkω0T
=−
e ( jkω T 1
− jkω0t
− jk ω0t0
t0
⎤⎦ − t0 − e jk ω0t0
)
0
2sin(kω0t0 ) = kω0T
2/11/2009 ECSE-303A
2t ⎞ ⎛ sin ⎜ π k 0 ⎟ T ⎠ ⎝ = πk
31
The ak’s are scaled samples of the continuous sinc function defined as
sin π x πx
sinc ( x )
This function equals to one at x=0 and has zeros at x=±n, n=1, 2, 3
sinc ( x )
1
-3
-2
-1
1
2
3
x 2/11/2009 ECSE-303A
32
The spectral coefficients then become
⎛ 2t ⎞ sin ⎜ π k 0 ⎟ T ⎠ ⎝ ak = {k ≠ 0} πk ⎡ ⎛ π k 2t0 ⎞ ⎤ sin 2t0 ⎢ ⎜⎝ T ⎟⎠ ⎥ 2t0 ⎛ k 2t0 ⎞ ⎢ ⎥ ak = sinc ⎜ = ⎟ k t 2 π T ⎢ 0 ⎥ T ⎝ T ⎠ ⎢⎣ ⎥⎦ T We define the duty cycle of the rectangular wave η
and therefore 2/11/2009 ECSE-303A
sinc ( x )
sin π x πx
2t0 T
ak = η sinc ( kη ) 33
For a 50% duty cycle, η=0.5
ak = η sinc ( kη ) 1 ⎛k⎞ = sinc ⎜ ⎟ 2 ⎝2⎠
-T/4
T/4 t
2t 1 0 a0 = ∫ 1dt = 0 T −t0 T
a0: is also taken into account in the sinc function
2/11/2009 ECSE-303A
34
Remember that k is a multiple of the fundamental frequency ω0, so for a 1 Hz (=2π rad/s) square wave, • a±1 represents the fundamental components at ω0=2π rad/sec, f0=1Hz, • a±2 represents the second harmonic components at ω0=4π rad/sec, f0=2Hz
2/11/2009 ECSE-303A
35
ak = η sinc ( kη ) For smaller values of duty cycle, the sinc envelope of the spectral coefficients expands, and ak coefficients are more densely packed under in each lobe of the sinc enveloppe.
2/11/2009 ECSE-303A
36
Example: Find the Fourier series coefficients of
x ( t ) = 2 + cos ω0t
x (t ) =
+∞
∑ae
k =−∞
ak =
jkω0 t
k
1 − jkω0 t x t e dt ( ) ∫ TT
Clearly here, because this function can be rewritten as
x (t ) = 2 + the Fourier coefficients are
a0 = 2 1 a1 = a−1 = 2
2/11/2009 ECSE-303A
e
jω0 t
+e 2
− jω0 t
For a real function x(t), a real coefficient ak involves a cosinus function
37
Example: Find the Fourier series coefficients of
x ( t ) = sin ω0t + 2sin 2ω0t
x (t ) =
+∞
∑ae
k =−∞
ak =
jkω0 t
k
1 − jkω0 t x t e dt ( ) ∫ TT
This function can be rewritten as
e jω0t − e − jω0t e j 2ω0t − e − j 2ω0t x (t ) = +2 2j 2j the Fourier coefficients are
1 a1 = −a−1 = 2j 1 a2 = −a−2 = j 2/11/2009 ECSE-303A
For a real function x(t), an imaginary coefficient ak involves a sinus function
38
Real form of the Fourier series If and only if the signal x(t) is real, x(t ) =
+∞
∑ae
k =−∞
jkω0t
k
+∞
(
= a0 + ∑ ak e jkω0t + a− k e− jkω0t k =1 +∞
)
(
= a0 + ∑ ak ⎡⎣cos ( kω0t ) + j sin ( kω0t ) ⎤⎦ + a− k ⎡⎣cos ( kω0t ) − j sin ( kω0t ) ⎤⎦ k =1
)
+∞
= a0 + ∑ ([ ak + a− k ] cos ( kω0t ) + [ ak − a− k ] j sin ( kω0t ) ) k =1
then we must have a-k=ak* in order to get ak+a-k and j(ak-a-k) to be real If we now represent the Fourier series coefficients as ak=Bk +jCk, we obtain the real form of the Fourier series +∞
x(t ) = a0 + ∑ ( Bk cos ( kω0t ) − Ck sin ( kω0t ) ) k =1
2/11/2009 ECSE-303A
39
Graph of the Fourier Series Coefficients: The Line Spectrum The set of complex Fourier series coefficients ak, {∞
LECTURE 12 Sections 3.1-3.3 Introduction to the Fourier series of periodic signals
Chapter 3: Fourier Series of periodic signals 3.0 3.1 3.2 3.3 3.4 3.5 3.8 3.9
Introduction Historical perspective The response of LTI systems to complex exponentials Fourier series representation of continuous-time periodic signals Convergence of Fourier series Properties of continuous-time Fourier series Fourier series and LTI systems Filtering
Q. How to make a rectangular shape out of circles?
2/11/2009 ECSE-303A
The Fourier series
By analogy, could we plot a square function using sinusoidal functions? f(t) t
In the 1800’s, Jean Baptiste Joseph Fourier found that most periodic functions with finite average power could be represented by a sum of sines and cosines
x(t ) =
+∞
∑
k =−∞
=
ak e jkω0t
+∞
∑a
k =−∞
k
⎡⎣ cos ( kω0t ) + j sin ( kω0t ) ⎤⎦
+∞
= a0 + ∑ Ak cos ( kω0t ) + Bk sin ( kω0t ) 2/11/2009 ECSE-303A
k =0
J.B. Fourier 1768-1830
Example
,
2/11/2009 ECSE-303A
f0 = 1Hz
+
f0 = 1Hz
f0 = 3Hz
= + f0 = 5Hz
2/11/2009 ECSE-303A
+ ... f0 = 7, 9...Hz
With three sin(2πf0t) components f0 = 1, 3, 5 Hz
With twentyfive sin(2πf0t) components f0 = 1, 3, 5...51 Hz 2/11/2009 ECSE-303A
To represent any arbitrary periodic function, what frequency component must be chosen? What is the amplitude coefficient of each term?
2/11/2009 ECSE-303A
Chapter 3: Fourier Series of periodic signals 3.0 Introduction 3.1 Historical perspective 3.2 The response of LTI systems to complex exponentials 3.3 Fourier series representation of continuous-time periodic signals 3.4 Convergence of Fourier series 3.5 Properties of continuous-time Fourier series 3.8 Fourier series and LTI systems 3.9 Filtering
Eigenfunctions of LTI Difference and Differential Systems We already know that the complex exponentials of the type Aest and Czn remain basically invariant under the action of time shifts (difference systems) and derivatives (differential systems). The response of an LTI system to a complex exponential input is the same complex exponential with only a change in (complex) amplitude. Continuous-time LTI system:
e → H ( s )e
Discrete-time LTI system:
z n → H ( z) z n
st
st
Where the complex amplitude factors H(s), H(z) are functions of the complex variable s or z.
2/11/2009 ECSE-303A
10
Input signals like x(t)=est and x[n]=zn for which the system output is a constant times the input signal are called eigenfunctions of the LTI system, and the complex gains are the system's eigenvalues corresponding to the eigenfunctions.
2/11/2009 ECSE-303A
11
To show that x(t)=est is indeed an eigenfunction of any LTI system of impulse response h(t), we look at the following convolution integral:
y (t ) =
+∞
∫ h (τ ) x ( t − τ ) dτ
−∞ +∞
=
∫
h (τ ) e s( t −τ ) dτ
−∞ +∞
= e st
∫
h (τ ) e − sτ dτ
−∞
The system's response has the form y(t)=H(s)est, where
z
+∞
H (s) = h(τ )e − sτ dτ is an eigenvalue and 2/11/2009 ECSE-303A
est
−∞
is an eigenfunction 12
Similarly for LTI discrete-time systems, the complex exponential x[n]=zn is an eigenfunction: +∞
y [ n] =
=
∑ h [k ] x [n − k ]
k =−∞ +∞
∑
k =−∞
= zn
h [ k ] z n−k
+∞
∑ h [k ] z
−k
k =−∞
The system's response has the form
y[n] = H ( z ) z n Where
H ( z) =
+∞
∑
h[k ]z − k
k =−∞
2/11/2009 ECSE-303A
13
The eigenvalues of LTI systems in continuous and discrete-time +∞
H ( s) =
∫
h(τ )e− sτ dτ
−∞
H ( z) =
+∞
∑ h[k ]z
−k
k =−∞
Are respectively known as the Laplace transform and the ztransform of the system's impulse response (also called transfer function, more of this later in the course and Signals 2).
2/11/2009 ECSE-303A
14
Chapter 3: Fourier Series of periodic signals 3.0 3.1 3.2 3.3 3.4 3.5 3.8 3.9
Introduction Historical perspective The response of LTI systems to complex exponentials Fourier series representation of continuous-time periodic signals Convergence of Fourier series Properties of continuous-time Fourier series Fourier series and LTI systems Filtering
Linear Combinations of HarmonicallyRelated Complex Exponentials Periodic signals satisfy
x (t ) = x (t + T )
{−∞ < t < +∞}
for some positive value of T. The smallest such T is the fundamental period and ω0=2π/T (radians/s) is the fundamental frequency A periodic signal x(t) is completely known from prior knowledge of only one period T A special case of periodic signals, the harmonically-related complex exponentials have frequencies that are integer multiples of ω0 jkω0t
φk (t ) e
2/11/2009 ECSE-303A
, k = 0,± 1,± 2,…
16
Harmonically related exponentials form an orthogonal set. That is, 2π
2π
ω0
ω0
0
0
* φ ( t ) φ ∫ k m (t )dt =
∫
e jkω0t e − jmω0t dt
2π
=
ω0
∫
e j ( k − m )ω0t dt
0
1 ⎡⎣e j ( k − m )2π − 1⎤⎦ = j (k − m)ω0 ⎧ 2π ⎪ , m=k = ⎨ ω0 ⎪ 0, m ≠ k ⎩ 2/11/2009 ECSE-303A
Use l’Hopital’s rule 17
A linear combination (i.e. summation) of harmonically-related complex exponentials φk(t)
x(t ) =
+∞
∑
k =−∞
akφk =
+∞
∑
k =−∞
ak e jkω0t =
+∞
∑ae
k =−∞
jk (
2π )t T
k
is also periodic with fundamental period T. Terms with k=±1 ⇒ The fundamental frequency component or the first harmonic of the signal Terms with k=±2 ⇒ second harmonic components (at frequency 2ω0) Terms with k=±N ⇒ Nth harmonic components.
The representation of periodic signals in the form given above is referred to as the Fourier series representation 2/11/2009 ECSE-303A
18
Example 4.1: Consider the periodic signal with fundamental frequency ω0=π/2 rad/s made up of the sum of five harmonic components:
x(t ) =
5
∑a e
k =−5
π
jk t 2
k
,
a0 = 0 a±1 = −0.2026 a±2 = 0 a±3 = −0.0225 a±4 = 0 a±5 = −0.0081 2/11/2009 ECSE-303A
19
Collecting the harmonic components together, we obtain
x(t ) =
5
∑a e
k =−5
π
jk t 2
k
= −0.2026(e
π
j t 2
+e
π
−j t 2
) − 0.0225(e
j
3π t 2
+e
−j
3π t 2
) − 0.0081(e
j
5π t 2
+e
−j
5π t 2
π
3π 5π = −0.4052 cos( t ) − 0.0450 cos( t ) − 0.0162 cos( t ) 2 2 2
2/11/2009 ECSE-303A
20
)
Relative amplitude
π 3π 5π x(t ) = −0.4052 cos( t ) − 0.0450 cos( t ) − 0.0162 cos( t ) 2 2 2
2/11/2009 ECSE-303A
Time [s]
21
Given a periodic function, how to represent it with a summation of harmonic components?
x(t ) =
+∞
∑ae
k =−∞
jkω0t
k
We use the orthogonality property of the harmonically related complex exponentials. Multiplying both sides of the above equation with another complex exponential and integrating over one fundamental period of the signal +∞
T
T
0
0 k =−∞
− jnω0t x ( t ) e dt = ∫ ∫
=
+∞
T
∑
∑ a ∫e
k =−∞
k
0
= Tak 2/11/2009 ECSE-303A
ak e jkω0t e − jnω0t dt T
jkω0t − jnω0 t
e
{k = n}
dt
1 − jnω0t ⇒ ak = ∫ x ( t ) e dt T 0 22
In conclusion, if a periodic signal x(t) has a Fourier series representation, then we have the Fourier series equation pair
x (t ) =
+∞
∑ae
k =−∞
k
jkω0t
=
+∞
∑ae
k =−∞
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
k
1 1 − jkω0t ak = ∫ x ( t ) e dt = ∫ x ( t ) e TT TT
2/11/2009 ECSE-303A
⎛ 2π − jk ⎜ ⎝ T
⎞ ⎟t ⎠
dt
23
x (t ) =
+∞
∑
k =−∞
ak e jkω0t =
+∞
∑ae
k =−∞
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
k
The first equation provides a time-domain representation of the signal as a sum of periodic complex exponential signals. This is the synthesis equation.
1 1 − jkω0t ak = ∫ x ( t ) e dt = ∫ x ( t ) e TT TT
⎛ 2π − jk ⎜ ⎝ T
⎞ ⎟t ⎠
dt
The second equation provides a frequency-domain representation of the signal. The ak represent the Fourier series coefficients, or the spectral coefficients of x(t). 2/11/2009 ECSE-303A
24
Example: The periodic “sawtooth” signal
The fundamental period is T=1s hence
ω0 =
2π = 2π rad/s T
The first ak coefficient with k=0.
1 1 − jkω0t ak = ∫ x( t )e dt ⇒ a0 = ∫ x( t )dt TT TT Since the average over one period is 0, we find a0=0. 2/11/2009 ECSE-303A
25
1 ak = ∫ x(t )e − jkω0t dt TT 1
= ∫ x(t )e − jk 2π t dt 0
1
= ∫ (1 − 2t )e − jk 2π t dt 0
− jk 2π t
1
− jk 2π t
1
⎡ ⎤ ⎡e ⎤ e = ⎢(1 − 2t ) −⎢ ( −2 ) ⎥ ⎥ − jk 2π ⎦ 0 ⎣ − jk 2π ⎦ 0 ⎣ 1 1 1 = + = jk 2π jk 2π jkπ −j = {k ≠ 0} kπ 2/11/2009 ECSE-303A
26
x (t ) =
+∞
∑ae
k =−∞
k = ±1
k
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
−j ⎧ ⎫ , k ≠ 0⎬ ⎨T = 1, ak = kπ ⎩ ⎭
k = ±1,2
k = ±1,2,3
2/11/2009 ECSE-303A
27
x (t ) =
+∞
∑ae
k =−∞
k
⎛ 2π jk ⎜ ⎝ T
⎞ ⎟t ⎠
−j⎫ ⎧ ⎨T = 1, ak = ⎬ kπ ⎭ ⎩
k = ±∞
2/11/2009 ECSE-303A
28
ECSE-303A Signals and Systems I Wednesday, February 11, 2009
LECTURE 13 Section 3.3-More examples of Fourier series Section 3.4-Convergence of the Fourier series
Example Consider the following periodic rectangular wave of 2π fundamental period T and fundamental frequency ω 0 =
T
.
2/11/2009 ECSE-303A
30
The Fourier series coefficients are, for k=0 t0
2t0 1 a0 = ∫ 1dt = T −t0 T
a0: average value of x(t)
x (t ) =
+∞
∑ae
k =−∞
ak =
jkω0 t
k
1 − jkω0 t x t e dt ( ) ∫ TT
and for k≠0 t
1 0 − jkω0t ak = ∫ e dt T − t0 1
=−
⎡⎣e jkω0T
=−
e ( jkω T 1
− jkω0t
− jk ω0t0
t0
⎤⎦ − t0 − e jk ω0t0
)
0
2sin(kω0t0 ) = kω0T
2/11/2009 ECSE-303A
2t ⎞ ⎛ sin ⎜ π k 0 ⎟ T ⎠ ⎝ = πk
31
The ak’s are scaled samples of the continuous sinc function defined as
sin π x πx
sinc ( x )
This function equals to one at x=0 and has zeros at x=±n, n=1, 2, 3
sinc ( x )
1
-3
-2
-1
1
2
3
x 2/11/2009 ECSE-303A
32
The spectral coefficients then become
⎛ 2t ⎞ sin ⎜ π k 0 ⎟ T ⎠ ⎝ ak = {k ≠ 0} πk ⎡ ⎛ π k 2t0 ⎞ ⎤ sin 2t0 ⎢ ⎜⎝ T ⎟⎠ ⎥ 2t0 ⎛ k 2t0 ⎞ ⎢ ⎥ ak = sinc ⎜ = ⎟ k t 2 π T ⎢ 0 ⎥ T ⎝ T ⎠ ⎢⎣ ⎥⎦ T We define the duty cycle of the rectangular wave η
and therefore 2/11/2009 ECSE-303A
sinc ( x )
sin π x πx
2t0 T
ak = η sinc ( kη ) 33
For a 50% duty cycle, η=0.5
ak = η sinc ( kη ) 1 ⎛k⎞ = sinc ⎜ ⎟ 2 ⎝2⎠
-T/4
T/4 t
2t 1 0 a0 = ∫ 1dt = 0 T −t0 T
a0: is also taken into account in the sinc function
2/11/2009 ECSE-303A
34
Remember that k is a multiple of the fundamental frequency ω0, so for a 1 Hz (=2π rad/s) square wave, • a±1 represents the fundamental components at ω0=2π rad/sec, f0=1Hz, • a±2 represents the second harmonic components at ω0=4π rad/sec, f0=2Hz
2/11/2009 ECSE-303A
35
ak = η sinc ( kη ) For smaller values of duty cycle, the sinc envelope of the spectral coefficients expands, and ak coefficients are more densely packed under in each lobe of the sinc enveloppe.
2/11/2009 ECSE-303A
36
Example: Find the Fourier series coefficients of
x ( t ) = 2 + cos ω0t
x (t ) =
+∞
∑ae
k =−∞
ak =
jkω0 t
k
1 − jkω0 t x t e dt ( ) ∫ TT
Clearly here, because this function can be rewritten as
x (t ) = 2 + the Fourier coefficients are
a0 = 2 1 a1 = a−1 = 2
2/11/2009 ECSE-303A
e
jω0 t
+e 2
− jω0 t
For a real function x(t), a real coefficient ak involves a cosinus function
37
Example: Find the Fourier series coefficients of
x ( t ) = sin ω0t + 2sin 2ω0t
x (t ) =
+∞
∑ae
k =−∞
ak =
jkω0 t
k
1 − jkω0 t x t e dt ( ) ∫ TT
This function can be rewritten as
e jω0t − e − jω0t e j 2ω0t − e − j 2ω0t x (t ) = +2 2j 2j the Fourier coefficients are
1 a1 = −a−1 = 2j 1 a2 = −a−2 = j 2/11/2009 ECSE-303A
For a real function x(t), an imaginary coefficient ak involves a sinus function
38
Real form of the Fourier series If and only if the signal x(t) is real, x(t ) =
+∞
∑ae
k =−∞
jkω0t
k
+∞
(
= a0 + ∑ ak e jkω0t + a− k e− jkω0t k =1 +∞
)
(
= a0 + ∑ ak ⎡⎣cos ( kω0t ) + j sin ( kω0t ) ⎤⎦ + a− k ⎡⎣cos ( kω0t ) − j sin ( kω0t ) ⎤⎦ k =1
)
+∞
= a0 + ∑ ([ ak + a− k ] cos ( kω0t ) + [ ak − a− k ] j sin ( kω0t ) ) k =1
then we must have a-k=ak* in order to get ak+a-k and j(ak-a-k) to be real If we now represent the Fourier series coefficients as ak=Bk +jCk, we obtain the real form of the Fourier series +∞
x(t ) = a0 + ∑ ( Bk cos ( kω0t ) − Ck sin ( kω0t ) ) k =1
2/11/2009 ECSE-303A
39
Graph of the Fourier Series Coefficients: The Line Spectrum The set of complex Fourier series coefficients ak, {∞
